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3 years ago

What is the missing term in the factorization? 100x^2−36=4(5x+?)(5x−3)

Mathematics

2 answers:

bogdanovich [222]3 years ago

6 0

To get the missing terms, we must completely factorize the Left Hand Side of the Equation.

We can do this in two different ways

<u>Method 1</u>

$100x^{2} -36= 4(25x^{2} -9)}$

$\Rightarrow 100x^{2} -36= 4((5x)^{2} -3^2)}$

Using difference of two squares we have

$\Rightarrow 100x^{2} -36= 4(5x+3)(5x-3)}$

By comparing to the given equation, the missing term is

$3$

<u>Method 2</u>

$100x^{2} -36$

$\Rightarrow 100x^{2} -36= (10x)^{2} -6^{2}$

Recall and apply the difference of two squares formula, to obtain,

$100x^{2} -36=(10x+6)(10x-6)$

We can still further factor 2 out of each factor to get,

$100x^{2} -36=2(5x+3) \times 2(5x-3)$

$\Rightarrow 100x^{2} -36=2\times2(5x+3)(5x-3)$

$\Rightarrow 100x^{2} -36=4(5x+3)(5x-3)$

By comparing to the given equation, the missing term is

$3$

LekaFEV [45]3 years ago

4 0

100x^2−36=4(5x+3)(5x−3)

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An archaeologist uses an accelerator mass spectrometer to find the age of a buried branch. At the 68\%68%68, percent confidence

Komok [63]

Answer:

B. 10000 years old, with a margin of error of 400 years

Step-by-step explanation:

Assuming this complete problem: "12. An archaeologist uses an accelerator mass spectrometer to find the age of a buried branch. At the 68% confidence level, the spectrometer estimates that the branch was 10,000 years old with a following could the spectrometer estimate as the age of the branch at the 95% confidence level?"

The possible options are:

A. 9500 years old, with a margin of error of 500 years

B. 10000 years old, with a margin of error of 400 years

C. 9500 years olds, with a margin of error of 50 years

D. 10000 years old, with a margin of error of 40 year

Solution to the problem

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The margin of error for this case is given by this formula:

$ME= z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$

For the first case the confidence was 68% so then $\alpha =1-0.68 =0.32$ and $\alpha/2 =0.16$ we can find a critical value on the normal standard distribution that accumulates 0.16 of the area on each tail and this value is $z=\pm 0.994$ , because P(z<-0.944) = 0.16 and P(Z>.944) = 0.16

The margin of error can be expressed like this:

$ME = z_{\alpha/2} SE$

We can solve for the standard error on this case and we got:

$SE = \frac{ME}{z_{\alpha/2}} =\frac{200}{0.994}=201.207$

And then for the new confidence interval we need to calculate the new $z_{\alpha}$ . For the first case the confidence was 95% so then $\alpha =1-0.95 =0.05$ and $\alpha/2 =0.025$ we can find a critical value on the normal standard distribution that accumulates 0.025 of the area on each tail and this value is $z=\pm 1.96$ , because P(z<-1.96) = 0.025 and P(Z>1.96) = 0.025. So then the new margin of error would be:

$ME = z_{\alpha/2} SE = 1.96*201.207=394.366$

The estimation for the mean not changes and from the before and new case is 1000 years. So then the best option for this case would be:

B. 10000 years old, with a margin of error of 400 years

And makes sense since a larger confidence interval means a wider interval.

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3 years ago

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Scorpion4ik [409]

Answer:
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Step-by-step explanation:

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2 years ago

A simple random sample from a population with a normal distribution of 99 body temperatures has xbar = 99.10°F and s = 0.64°F. C

motikmotik

The general formula for the margin of error is e = (zs)/√n, when the sample size is 30 or more (otherwise, we'd need to to a t-interval).

Since we're needing a 99% confidence interval, we need to know what the positive z-score associated with this two-tailed area under the normal curve is. This can be a little tricky if you're using a standard normal table. What you want is a two-tailed interval that has an area of .99. What this means is that the remaining 0.01 is divided into 0.005 on each end. This then means that, to the right of the mean (which is 0), the area is 0.005 less than the total right-half area of 0.5, or 0.495. The total cumulative area, then, includes this plus the left half of .5, or 0.5 + 0.495 = .0995.

Then, if all we have to go on is a cumulative standard normal table, then we need to find the area closest to 0.995 and find the corresponding z-score. This z-score is 2.58.

Now that we have the z-score, we can now plug in all the values into the formula.

e = (2.58 · 0.66)/√102 = 0.1686.

So the 99% confidence interval will be that far above and below the mean (which is 98.8). So the lower limit is 98.8 - 0.1686 = 98.631, and the upper limit is 98.8 + 0.1686 = 98.967.

The TI-84 method:

This is much easier, if you're allowed to use this technology. You begin with the STAT button, and then move over to the TESTS menu. Then select ZInterval. We have summary stats rather than data in this problem, so we select Stats as the input. For the standard deviation, we put 0.66, for x-bar 98.8, for n 102, and for C-level 0.99. Then select Calculate, and it will provide the interval at the top of the screen (98.632, 98.968). Note that the lower limit is slightly different by this method. This is because, when doing it by hand, we had to approximate the z-score. This created a rounding error, albeit a small one.

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3 years ago

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GREYUIT [131]

Answer:

I'd say A is the correct answer. There's a chance I'm wrong but I don't think so

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2 years ago

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(2x^2-4x+8)+(3x^2+15x-12)

Neko [114]

Answer:

5x2+11x−4

Step-by-step explanation:

=2x2+−4x+8+3x2+15x+−12

Combine Like Terms:

=2x2+−4x+8+3x2+15x+−12

=(2x2+3x2)+(−4x+15x)+(8+−12)

=5x2+11x+−4

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2 years ago

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